IB exam questions: topic 2

1. We express the right-hand side as a power of 3: $\sqrt{3}=3^{1/2}$, so ${\left(\sqrt{3}\right)}^{126}=3^{126/2}=3^{63}$.
2. The equation is $3^{x^2-1}=3^{63}$. Since the exponential function is one-to-one, we equate the exponents: $x^2-1=63$.
3. Then $x^2=64$, so $x=\pm 8$.
4. Both values are valid (there are no domain restrictions).

Answer: $x=8$ or $x=-8$

1. We take $\ln$ on both sides: $(x-1)\ln 8=3x\ln 6$.
2. We write $\ln 8=3\ln 2$ and $\ln 6=\ln 2+\ln 3$: $3(x-1)\ln 2=3x(\ln 2+\ln 3)$.
3. We divide by 3: $x\ln 2-\ln 2=x\ln 2+x\ln 3$.
4. The terms $x\ln 2$ cancel: $-\ln 2=x\ln 3$.
5. We solve for $x$.

Answer: $x=-\frac{\ln 2}{\ln 3}$

1. We take $\ln$ on both sides: $x\ln 3+(2x+1)\ln 4=(x+2)\ln 6$.
2. We use $\ln 4=2\ln 2$ and $\ln 6=\ln 2+\ln 3$: $$x\ln 3+2(2x+1)\ln 2=(x+2)(\ln 2+\ln 3)$$
3. We expand: $x\ln 3+4x\ln 2+2\ln 2=x\ln 2+x\ln 3+2\ln 2+2\ln 3$.
4. We cancel $x\ln 3$ and $2\ln 2$ on both sides: $4x\ln 2=x\ln 2+2\ln 3$.
5. Then $3x\ln 2=2\ln 3$, so $x=\frac{2\ln 3}{3\ln 2}=\frac{\ln 3^2}{\ln 2^3}$.

Answer: $x=\frac{\ln 9}{\ln 8}$

1. We write $5^{x+1}=5\cdot 5^x$ and let $u=5^x>0$. The equation is $10u=1+\frac{3}{u}$.
2. We multiply by $u$: $10u^2=u+3$, that is, $10u^2-u-3=0$.
3. General formula: $u=\frac{1\pm \sqrt{1+120}}{20}=\frac{1\pm 11}{20}$, so $u=\frac{3}{5}$ or $u=-\frac{1}{2}$.
4. We reject $u=-\frac{1}{2}$, because $5^x>0$.
5. From $5^x=\frac{3}{5}$ we get $x={\log }_5\frac{3}{5}={\log }_{5}3-{\log }_{5}5=-1+{\log }_{5}3$.

Answer: $x=-1+{\log }_{5}3$

1. We rewrite using the laws of exponents: $2^{2x+3}=2^3\cdot (2^x{)}^2=8(2^x{)}^2$ and $2^{x+1}=2\cdot 2^x$.
2. Let $y=2^x>0$. The equation is $8y^2=2y+3$, that is, $8y^2-2y-3=0$.
3. General formula: $y=\frac{2\pm \sqrt{4+96}}{16}=\frac{2\pm 10}{16}$, so $y=\frac{3}{4}$ or $y=-\frac{1}{2}$.
4. We reject $y=-\frac{1}{2}$, because $2^x>0$ always.
5. With $y=\frac{3}{4}$: $2^x=\frac{3}{4}$, so $x={\log }_2\frac{3}{4}={\log }_{2}3-{\log }_{2}4=-2+{\log }_{2}3$.
6. Comparing with $a+{\log }_{2}b$: $a=-2$ and $b=3$.

Answer: $a=-2$, $b=3$

a) 1. Let $y=4^x>0$. Then $4^{-x}=\frac{1}{y}$ and the equation is $2y+\frac{1}{y}=3$. 2. Multiply by $y$: $2y^2+1=3y$, that is, $2y^2-3y+1=0$. 3. Factorise: $(2y-1)(y-1)=0$, so $y=\frac{1}{2}$ or $y=1$. Both are positive. 4. If $4^x=\frac{1}{2}$: $2^{2x}=2^{-1}$, so $2x=-1$ and $x=-\frac{1}{2}$. 5. If $4^x=1$: $x=0$.

b) (i)

1. Take natural logarithms: $x\ln a=2x+1$.
2. Group the terms in $x$: $x(\ln a-2)=1$.
3. If $\ln a\ne 2$, solve for $x$: $x=\frac{1}{\ln a-2}$.

b) (ii)

1. If $\ln a-2=0$, the equation $x\cdot 0=1$ is impossible, so there is no solution.
2. $\ln a=2$ is equivalent to $a=e^2$. (Check: $(e^2{)}^x=e^{2x}\ne e^{2x+1}$ for all $x$.)

Answer: a) $x=-\frac{1}{2}$ or $x=0$; b) (i) $x=\frac{1}{\ln a-2}$; (ii) $a=e^2$

1. We write everything in base 2: $4^{x-1}=\frac{(2^x{)}^2}{4}$.
2. Let $y=2^x>0$. The equation is $\frac{y^2}{4}=y+8$.
3. Multiply by 4: $y^2=4y+32$, that is, $y^2-4y-32=0$.
4. Factorise: $(y-8)(y+4)=0$, so $y=8$ or $y=-4$.
5. We reject $y=-4$, because $2^x>0$.
6. With $y=8$: $2^x=2^3$, so $x=3$.
7. Check: $4^2=16$ and $2^3+8=16$.

Answer: $x=3$

1. We write everything in base 2: $4^x=(2^x{)}^2$ and $2^{x+2}=2^x\cdot 2^2=4\cdot 2^x$.
2. Substitution $t=2^x$, with $t>0$ (an exponential is always positive): $t^2+4t=3$, that is, $t^2+4t-3=0$.
3. Quadratic formula: $$t=\frac{-4\pm \sqrt{16+12}}{2}=\frac{-4\pm \sqrt{28}}{2}=-2\pm \sqrt{7}.$$
4. We reject $t=-2-\sqrt{7}<0$. We are left with $t=\sqrt{7}-2\approx 0.646>0$, which is valid.
5. Undo the substitution: $2^x=\sqrt{7}-2$, so $x={\log }_2\left(\sqrt{7}-2\right)$.

Answer: $x={\log }_2\left(\sqrt{7}-2\right)\approx -0.631$

1. We factorise each argument: $x^2-1=(x-1)(x+1)$ and $x^2+x=x(x+1)$.
2. We move the coefficient 2 into the exponent: $2\ln (x+1)=\ln (x+1{)}^2$.
3. We use $\ln a+\ln b=\ln (ab)$ and $\ln a-\ln b=\ln \frac{a}{b}$: $$\ln \frac{(x-1)(x+1)\cdot x(x+1)}{(x+1{)}^2}$$
4. We cancel the factor $(x+1{)}^2$ from the numerator and the denominator: $$\ln (x(x-1))=\ln (x^2-x)$$

Answer: $\ln (x^2-x)$

1. Let $t={\log }_{x}y$. By the change of base formula for inverses, ${\log }_{y}x=\frac{1}{{\log }_{x}y}=\frac{1}{t}$.
2. The equation becomes $t=\frac{4}{t}$, that is, $t^2=4$, so $t=2$ or $t=-2$.
3. If ${\log }_{x}y=2$, then by definition of logarithm $y=x^2$.
4. If ${\log }_{x}y=-2$, then $y=x^{-2}=\frac{1}{x^2}$.
5. Conditions: $x>0$, $x\ne 1$, $y>0$, $y\ne 1$; both expressions satisfy these for $x>0$, $x\ne 1$.

Answer: $y=x^2$ or $y=\frac{1}{x^2}$

a) 1. Change of base to base 2: ${\log }_4{x}^3=\frac{{\log }_2{x}^3}{{\log }_{2}4}=\frac{3{\log }_{2}x}{2}$. 2. ${\log }_2{x}^2=2{\log }_{2}x$. 3. ${\log }_{8}128^3=\frac{{\log }_{2}128^3}{{\log }_{2}8}=\frac{3\cdot 7}{3}=7$, because $128=2^7$ and $8=2^3$. 4. The equation in base 2 is $$\frac{{\log }_2{x}^3}{{\log }_{2}4}+{\log }_2{x}^2=\frac{{\log }_{2}128^3}{{\log }_{2}8},$$ which, simplified, is $\frac{3}{2}{\log }_{2}x+2{\log }_{2}x=7$.

b) 1. We add the terms in ${\log }_{2}x$: $\frac{7}{2}{\log }_{2}x=7$. 2. We solve: ${\log }_{2}x=2$. 3. By definition, $x=2^2=4$. It is valid because $x>0$.

Answer: a) $\frac{{\log }_2{x}^3}{{\log }_{2}4}+{\log }_2{x}^2=\frac{{\log }_{2}128^3}{{\log }_{2}8}$; b) $x=4$

1. We apply $\log p+\log q=\log pq$ and $\frac{1}{2}\log A=\log \sqrt{A}$ to the right-hand side: $\frac{1}{2}({\log }_{10}p+{\log }_{10}q)={\log }_{10}\sqrt{pq}$.
2. Equate the arguments: $\frac{p+2q}{2\sqrt{2}}=\sqrt{pq}$.
3. Multiply by $2\sqrt{2}$: $p+2q=2\sqrt{2}\sqrt{pq}$.
4. Square both sides (both are positive): $p^2+4pq+4q^2=8pq$.
5. Move everything to one side: $p^2-4pq+4q^2=0$, that is, $(p-2q{)}^2=0$.
6. Therefore $p=2q$.
7. Check: $\frac{2q+2q}{2\sqrt{2}}=\sqrt{2}q$ and $\sqrt{2q\cdot q}=\sqrt{2}q$. They coincide.

Answer: $p=2q$

1. Change of base to base 3: ${\log }_{9}6=\frac{{\log }_{3}6}{{\log }_{3}9}=\frac{{\log }_{3}6}{2}$.
2. Then $10{\log }_{9}6=5{\log }_{3}6=5{\log }_3(3\cdot 2)=5(1+{\log }_{3}2)=5+5{\log }_{3}2$.
3. The equation is $m-n{\log }_{3}2=5+5{\log }_{3}2$.
4. Since ${\log }_{3}2$ is irrational and $m,n$ are integers, the integer part and the coefficient of ${\log }_{3}2$ must match: $m=5$ and $-n=5$, that is, $n=-5$.

Answer: $m=5,n=-5$

a) By the change of base to base 3: $${\log }_9{x}^3=\frac{{\log }_3{x}^3}{{\log }_{3}9}=\frac{3{\log }_{3}x}{2}=\frac{3}{2}{\log }_{3}x,\text{so }k=\frac{3}{2}.$$ Similarly, ${\log }_{27}512=\frac{{\log }_{3}512}{{\log }_{3}27}=\frac{{\log }_{3}512}{3}$. Since $512=8^3$, ${\log }_{3}512=3{\log }_{3}8$, and therefore ${\log }_{27}512={\log }_{3}8$. Thus $m=1$.

b) Domain condition: $x>0$.

1. Using part (a): ${\log }_9{x}^3=\frac{3}{2}{\log }_{3}x$ and ${\log }_3{x}^{1/2}=\frac{1}{2}{\log }_{3}x$. Also ${\log }_{27}512={\log }_{3}8$.
2. The equation becomes $\frac{3}{2}{\log }_{3}x+\frac{1}{2}{\log }_{3}x={\log }_{3}8$, that is, $2{\log }_{3}x={\log }_{3}8$.
3. Then ${\log }_3{x}^2={\log }_{3}8$, so $x^2=8$ and $x=\pm 2\sqrt{2}$.
4. We reject $x=-2\sqrt{2}$ because $x>0$. This leaves $x=2\sqrt{2}$.

Answer: a) $k=\frac{3}{2}$, $m=1$; b) $x=2\sqrt{2}$

1. We change each logarithm to base 2 using ${\log }_{a}b=\frac{{\log }_{2}b}{{\log }_{2}a}$: $$a=\frac{{\log }_{2}3}{{\log }_{2}2}\cdot \frac{{\log }_{2}4}{{\log }_{2}3}\cdot \frac{{\log }_{2}5}{{\log }_{2}4}\cdots \frac{{\log }_{2}32}{{\log }_{2}31}.$$
2. The product is telescopic: each numerator cancels with the next denominator, leaving only ${\log }_{2}32$ on top and ${\log }_{2}2$ on the bottom.
3. $a=\frac{{\log }_{2}32}{{\log }_{2}2}={\log }_{2}32={\log }_2{2}^5=5$, which is an integer, as the question states.

Answer: $a={\log }_{2}32=5$

1. We write it in exponential form: ${\log }_{16}A=\frac{1}{2}⟺A=16^{1/2}=4$. Here $A=\sqrt[3]{100-x^2}$, so $\sqrt[3]{100-x^2}=4$.
2. We cube both sides: $100-x^2=64$.
3. We solve: $x^2=36$, so $x=\pm 6$.
4. Check: the argument of the logarithm must be positive; with $x=\pm 6$ it equals $100-36=64>0$, so both solutions are valid.

Answer: $x=6$ or $x=-6$

1. By definition of absolute value, $|\ln (x+3)|=1$ is equivalent to $\ln (x+3)=1$ or $\ln (x+3)=-1$.
2. Case $\ln (x+3)=1$: $x+3=e$, so $x=e-3$.
3. Case $\ln (x+3)=-1$: $x+3=e^{-1}$, so $x=e^{-1}-3$.
4. Domain: we need $x+3>0$. Since $e-3\approx -0.28$ and $e^{-1}-3\approx -2.63$, both are greater than $-3$, so both solutions are valid.

Answer: $x=e-3$ or $x=e^{-1}-3$

1. We bring the logarithms to the same side: ${\log }_{2}x-{\log }_{2}5={\log }_2\frac{x}{5}$.
2. We write the right-hand side as a single logarithm: $2={\log }_{2}4$, so $2+{\log }_{2}3={\log }_{2}4+{\log }_{2}3={\log }_{2}12$.
3. Equate the arguments: $\frac{x}{5}=12$, so $x=60$.
4. It is valid because $x>0$.

Answer: $x=60$

1. We use $\log a+\log b=\log ab$: ${\log }_2((x+3)(x-3))=4$.
2. By definition of logarithm: $(x+3)(x-3)=2^4=16$.
3. Difference of squares: $x^2-9=16$, so $x^2=25$ and $x=5$ or $x=-5$.
4. Domain: $x+3>0$ and $x-3>0$, that is, $x>3$. We reject $x=-5$ (it would give logarithms of negative numbers).
5. Check with $x=5$: ${\log }_{2}8+{\log }_{2}2=3+1=4$.

Answer: $x=5$

a) 1. We bring the exponents out in front: $\ln 2^{4x-1}=(4x-1)\ln 2$ and $\ln 8^{x+5}=\ln 2^{3(x+5)}=3(x+5)\ln 2$. 2. The last term is a base-2 logarithm of a power of 2: ${\log }_{2}16^{1-2x}=(1-2x){\log }_{2}16=4(1-2x)$. 3. The equation becomes: $$(4x-1)\ln 2=3(x+5)\ln 2+4(1-2x)$$ 4. We expand: $4x\ln 2-\ln 2=3x\ln 2+15\ln 2+4-8x$. 5. We group the terms in $x$ on the left: $4x\ln 2-3x\ln 2+8x=16\ln 2+4$. 6. We factor out $x$: $x(\ln 2+8)=16\ln 2+4$, and solve for $x$: $$x=\frac{4+16\ln 2}{8+\ln 2}\approx 1.7359$$

b) 1. From ${\log }_{a}x=2$ we get $a^2=x$, that is, $a=\sqrt{x}$ (the base must be positive, so $-\sqrt{x}$ is rejected). 2. $a=\sqrt{1.7359}\approx 1.3175$, which to three decimal places is $1.318$.

Answer: a) $x=\frac{4+16\ln 2}{8+\ln 2}$ b) $a=1.318$

a) 1. Domain: $x-2>0$, that is, $x>2$ (and $x^2-6x+12=(x-3{)}^2+3>0$ always). 2. We change ${\log }_4$ to base 2: ${\log }_{4}A=\frac{{\log }_{2}A}{{\log }_{2}4}=\frac{1}{2}{\log }_{2}A$. 3. The equation is ${\log }_2(x-2)=\frac{1}{2}{\log }_2(x^2-6x+12)$; multiplying by 2: ${\log }_2(x-2{)}^2={\log }_2(x^2-6x+12)$. 4. We equate the arguments: $x^2-4x+4=x^2-6x+12$. 5. Then $2x=8$, so $x=4$. 6. Check: $4>2$ satisfies the domain; ${\log }_{2}2=1$ and ${\log }_4(16-24+12)={\log }_{4}4=1$.

b) 1. Domain: $x>0$. 2. We take $\ln$ on both sides: $\ln x\cdot \ln x=(\ln x{)}^3$, that is, $(\ln x{)}^2=(\ln x{)}^3$. 3. We move everything to one side and factorise: $(\ln x{)}^2(\ln x-1)=0$. 4. So $\ln x=0$ or $\ln x=1$, hence $x=1$ or $x=e$. Both are positive, so both are valid.

Answer: a) $x=4$ b) $x=1x=e$

1. Domain: $x-3>0$ and $x+1>0$, that is, $x>3$.
2. We change to base 3: ${\log }_{1/3}(x+1)=\frac{{\log }_3(x+1)}{{\log }_3\frac{1}{3}}=-{\log }_3(x+1)$.
3. Also, $2{\log }_3(x-3)={\log }_3(x-3{)}^2$. The equation is: $${\log }_3\frac{(x-3{)}^2}{x+1}=2$$
4. We write it in exponential form: $\frac{(x-3{)}^2}{x+1}=3^2=9$.
5. Then $x^2-6x+9=9x+9$, that is, $x^2-15x=0$, so $x(x-15)=0$.
6. Solutions: $x=0$ or $x=15$. We reject $x=0$ because it does not satisfy $x>3$.
7. Check with $x=15$: $2{\log }_{3}12-{\log }_{3}16={\log }_3\frac{144}{16}={\log }_{3}9=2$.

Answer: $x=15$

1. Domain: $x>0$ and $x\ne 1$.
2. We use the change of base: ${\log }_{x}5=\frac{1}{{\log }_{5}x}$. Let $t={\log }_{5}x$ (with $t\ne 0$).
3. The equation becomes $9t=\frac{25}{t}$, that is, $9t^2=25$.
4. Then $t^2=\frac{25}{9}$ and $t=\pm \frac{5}{3}$.
5. We undo the substitution: $x=5^t$, so $x=5^{5/3}$ or $x=5^{-5/3}$.
6. Both values are positive and different from 1, so they satisfy the domain.

Answer: $x=5^{\frac{5}{3}}$ or $x=5^{-\frac{5}{3}}$

1. Domain: $x>0$. Let $u=\ln x$.
2. The equation is $2u^2=3u-1$, that is, $2u^2-3u+1=0$.
3. Factorise: $(2u-1)(u-1)=0$, so $u=\frac{1}{2}$ or $u=1$.
4. If $\ln x=\frac{1}{2}$, then $x=e^{1/2}=\sqrt{e}$.
5. If $\ln x=1$, then $x=e$.
6. Both values are positive, so they are valid.

Answer: $x=\sqrt{e}$ or $x=e$

1. Domain: $x+7>0$ and $2x>0$, that is, $x>0$.
2. Change of base: ${\log }_{1/3}2x=\frac{{\log }_{3}2x}{{\log }_3(1/3)}=-{\log }_{3}2x$.
3. The equation becomes $2-{\log }_3(x+7)=-{\log }_3(2x)$, that is, $2={\log }_3(x+7)-{\log }_3(2x)$.
4. Quotient property: ${\log }_3\frac{x+7}{2x}=2$.
5. Exponential form: $\frac{x+7}{2x}=3^2=9$.
6. Solve for $x$: $x+7=18x$, so $17x=7$ and $x=\frac{7}{17}$.
7. It satisfies $x>0$, so it is valid.

Answer: $x=\frac{7}{17}$

Condition: $x>0$.

1. We use ${\log }_3\sqrt{x}=\frac{1}{2}{\log }_{3}x$.
2. Change of base: $\frac{1}{{\log }_{2}3}={\log }_{3}2$, so $\frac{1}{2{\log }_{2}3}=\frac{1}{2}{\log }_{3}2$.
3. ${\log }_3(4x^3)={\log }_{3}4+3{\log }_{3}x=2{\log }_{3}2+3{\log }_{3}x$.
4. The equation becomes $\frac{1}{2}{\log }_{3}x=\frac{1}{2}{\log }_{3}2+2{\log }_{3}2+3{\log }_{3}x=\frac{5}{2}{\log }_{3}2+3{\log }_{3}x$.
5. Grouping terms: $-\frac{5}{2}{\log }_{3}x=\frac{5}{2}{\log }_{3}2$, so ${\log }_{3}x=-{\log }_{3}2={\log }_3\frac{1}{2}$.
6. Therefore $x=\frac{1}{2}$, which satisfies $x>0$.

Answer: $x=\frac{1}{2}$

1. From the first equation, $\ln \frac{x}{y}=1$ implies $\ln x-\ln y=1$, so $\ln x=1+\ln y$.
2. In the second, we use $\ln x^3=3\ln x$ and $\ln y^2=2\ln y$: $3\ln x+2\ln y=5$.
3. We substitute for $\ln x$: $3(1+\ln y)+2\ln y=5$, that is, $3+5\ln y=5$.
4. We solve: $\ln y=\frac{2}{5}$, so $y=e^{2/5}$.
5. Then $\ln x=1+\frac{2}{5}=\frac{7}{5}$, so $x=e^{7/5}$.
6. Check: $x,y>0$ (the exponential is always positive), and $\frac{x}{y}=e^{7/5-2/5}=e$, so $\ln \frac{x}{y}=1$.

Answer: $x=e^{\frac{7}{5}}y=e^{\frac{2}{5}}$

1. The first equation $\ln \frac{y}{x}=2$ is equivalent to $\frac{y}{x}=e^2$, so $y=e^{2}x$ (with $x,y>0$).
2. Second equation, using logarithm properties: $2\ln x+3\ln y=7$.
3. From $y=e^{2}x$ we have $\ln y=2+\ln x$. Substitute: $2\ln x+3(2+\ln x)=7$.
4. Expand: $5\ln x+6=7$, so $\ln x=\frac{1}{5}$ and $x=e^{1/5}$.
5. Then $y=e^2\cdot e^{1/5}=e^{11/5}$.
6. Both are positive, so they satisfy the domain.

Answer: $x=e^{1/5},y=e^{11/5}$

1. We write both sides of the second equation with base 2: $4^x=2^{2x}$ and $8^y=2^{3y}$.
2. Equating exponents: $2x=3y$, i.e. $x=\frac{3y}{2}$.
3. Substitute into the first equation: $\frac{3y}{2}+2y=5$, i.e. $\frac{7y}{2}=5$, so $y=\frac{10}{7}$.
4. Hence $x=\frac{3}{2}\cdot \frac{10}{7}=\frac{15}{7}$.
5. Check: $x+2y=\frac{15}{7}+\frac{20}{7}=5$ and $2x=\frac{30}{7}=3y$.

Answer: $x=\frac{15}{7},y=\frac{10}{7}$

1. First equation: ${\log }_{2}6x=1+2{\log }_{2}y$. Since $1={\log }_{2}2$ and $2{\log }_{2}y={\log }_2{y}^2$, we get ${\log }_{2}6x={\log }_{2}2y^2$, so $6x=2y^2$.
2. Second equation: $1={\log }_{6}6$, so ${\log }_{6}6x={\log }_6(15y-25)$ and therefore $6x=15y-25$.
3. Equate the two expressions for $6x$: $2y^2=15y-25$, that is, $2y^2-15y+25=0$.
4. Quadratic formula: $y=\frac{15\pm \sqrt{225-200}}{4}=\frac{15\pm 5}{4}$, so $y=5$ or $y=\frac{5}{2}$.
5. From $6x=2y^2$ we get $x=\frac{y^2}{3}$. If $y=5$: $x=\frac{25}{3}$. If $y=\frac{5}{2}$: $x=\frac{25/4}{3}=\frac{25}{12}$.
6. Domain check: $x>0$, $y>0$ and $15y-25>0$. For $y=5$: $50>0$. For $y=\frac{5}{2}$: $37.5-25=12.5>0$. Both solutions are valid.

Answer: $x=\frac{25}{3},y=5$ and $x=\frac{25}{12},y=\frac{5}{2}$

Domain conditions: $x+1>0$, $x+1\ne 1$, $y>0$, $y+1>0$, $y+1\ne 1$.

1. From the first equation, by the definition of a logarithm: $y=(x+1{)}^2$.
2. From the second: $x=(y+1{)}^{1/4}$, i.e. $x^4=y+1$ (with $x>0$).
3. Substitute for $y$: $x^4=(x+1{)}^2+1=x^2+2x+2$, so $x^4-x^2-2x-2=0$.
4. Try $x=-1$: $1-1+2-2=0$, so it is a root, and factorising gives $(x+1)(x^3-x^2-2)=0$.
5. $x=-1$ is not valid (it would give base $x+1=0$). This leaves $x^3-x^2-2=0$.
6. Let $f(x)=x^3-x^2-2$. Since $f(1)=-2<0$ and $f(2)=2>0$, there is a root between 1 and 2. It is the only real root: $f^{\prime }(x)=3x^2-2x$ is positive for $x>\frac{2}{3}$ ($f$ increasing), and for $x\le \frac{2}{3}$ we have $f(x)\le f(0)=-2<0$. Using the GDC, $x\approx 1.6956$.
7. Hence $y=(x+1{)}^2\approx (2.6956{)}^2\approx 7.27$.
8. Check: $x^4\approx 8.27=y+1$, and the bases $x+1\approx 2.70$ and $y+1\approx 8.27$ are positive and different from 1.