Question

If 3^x =4^x-1 ,then find x

Answer 1

Step-by-step explanation:

3^x = 4^x-1

3^x = 4^x / 4

of 4 * 3^x = 4^x

To solve this, let us take log on both sides. Taking log on both sides, we get

log 4 + x log 3  = x log 4

or x (log4-log3) = log 4

or x = log 4 / (log 4 - log 3)

log 4 = 0.6021 and log 3 = 0.4771

Thus, we get x = 0.6021 / (0.6021-0.4771) = 0.6021/0.125 = 4.8168

Thus, x = 4.8168.

Answer 2

Answer:

$3^x=4^{x-1}\quad :\quad x=\frac{2\ln \left(2\right)}{\ln \left(\frac{4}{3}\right)}\quad \left(\mathrm{Decimal}:\quad x=4.81884\dots \right)$

Step-by-step explanation:

$3^x=4^{x-1}$

$\mathrm{If\:}f\left(x\right)=g\left(x\right)\mathrm{,\:then\:}\ln \left(f\left(x\right)\right)=\ln \left(g\left(x\right)\right)$

$\ln \left(3^x\right)=\ln \left(4^{x-1}\right)$

$\mathrm{Apply\:log\:rule}:\quad \log _a\left(x^b\right)=b\cdot \log _a\left(x\right)$

$\ln \left(3^x\right)=x\ln \left(3\right),\:\space\ln \left(4^{x-1}\right)=\left(x-1\right)\ln \left(4\right)$

$x\ln \left(3\right)=\left(x-1\right)\ln \left(4\right)$

$\mathrm{Simplify\:}\left(x-1\right)\ln \left(4\right)$

$\left(x-1\right)\ln \left(4\right)$

$\mathrm{Simplify}\:\ln \left(4\right)$

$\ln \left(4\right)$

$\mathrm{Rewrite\:}4\mathrm{\:in\:power-base\:form:}\quad 4=2^2$

$=\ln \left(2^2\right)$

$\mathrm{Apply\:log\:rule}:\quad \log _a\left(x^b\right)=b\cdot \log _a\left(x\right)$

$\ln \left(2^2\right)=2\ln \left(2\right)$

$=2\ln \left(2\right)$

$=2\ln \left(2\right)\left(x-1\right)$

$\mathrm{Solve\:}\:x\ln \left(3\right)=2\ln \left(2\right)\left(x-1\right)$

$x\ln \left(3\right)=2\ln \left(2\right)\left(x-1\right)$

$\mathrm{Expand\:}2\ln \left(2\right)\left(x-1\right)$

$2\ln \left(2\right)\left(x-1\right)$

$\mathrm{Apply\:the\:distributive\:law}:\quad \:a\left(b-c\right)=ab-ac$

$a=2\ln \left(2\right),\:b=x,\:c=1$

$=2\ln \left(2\right)x-2\ln \left(2\right)\cdot \:1$

$=2\ln \left(2\right)x-2\cdot \:1\cdot \ln \left(2\right)$

$\mathrm{Multiply\:the\:numbers:}\:2\cdot \:1=2$

$=2\ln \left(2\right)x-2\ln \left(2\right)$

$x\ln \left(3\right)=2\ln \left(2\right)x-2\ln \left(2\right)$

$\mathrm{Subtract\:}2\ln \left(2\right)x\mathrm{\:from\:both\:sides}$

$x\ln \left(3\right)-2\ln \left(2\right)x=2\ln \left(2\right)x-2\ln \left(2\right)-2\ln \left(2\right)x$

$\mathrm{Simplify}$

$x\ln \left(3\right)-2\ln \left(2\right)x=-2\ln \left(2\right)$

$\mathrm{Factor}\:x\ln \left(3\right)-2\ln \left(2\right)x$

$x\ln \left(3\right)-2\ln \left(2\right)x$

$\mathrm{Factor\:out\:common\:term\:}x$

$=x\left(\ln \left(3\right)-2\ln \left(2\right)\right)$

$\left(\ln \left(3\right)-2\ln \left(2\right)\right)x=-2\ln \left(2\right)$

$\mathrm{Divide\:both\:sides\:by\:}\ln \left(3\right)-2\ln \left(2\right)$

$\dfrac{\left(\ln \left(3\right)-2\ln \left(2\right)\right)x}{\ln \left(3\right)-2\ln \left(2\right)}=\dfrac{-2\ln \left(2\right)}{\ln \left(3\right)-2\ln \left(2\right)}$

$\mathrm{Simplify}$

$\dfrac{\left(\ln \left(3\right)-2\ln \left(2\right)\right)x}{\ln \left(3\right)-2\ln \left(2\right)}=\dfrac{-2\ln \left(2\right)}{\ln \left(3\right)-2\ln \left(2\right)}$

$\mathrm{Simplify\:}\dfrac{\left(\ln \left(3\right)-2\ln \left(2\right)\right)x}{\ln \left(3\right)-2\ln \left(2\right)}$

$\dfrac{\left(\ln \left(3\right)-2\ln \left(2\right)\right)x}{\ln \left(3\right)-2\ln \left(2\right)}$

$\mathrm{Cancel\:the\:common\:factor:}\:\ln \left(3\right)-2\ln \left(2\right)$

$=x$

$\mathrm{Simplify\:}\dfrac{-2\ln \left(2\right)}{\ln \left(3\right)-2\ln \left(2\right)}$

$\dfrac{-2\ln \left(2\right)}{\ln \left(3\right)-2\ln \left(2\right)}$

$\mathrm{Apply\:the\:fraction\:rule}:\quad \dfrac{-a}{-b}=\dfrac{a}{b}$

$\ln \left(3\right)-2\ln \left(2\right)=-\left(2\ln \left(2\right)-\ln \left(3\right)\right)$

$=\dfrac{2\ln \left(2\right)}{2\ln \left(2\right)-\ln \left(3\right)}$

$2\ln \left(2\right)-\ln \left(3\right)$

$\mathrm{Apply\:log\:rule}:\quad \:a\log _c\left(b\right)=\log _c\left(b^a\right)$

$2\ln \left(2\right)=\ln \left(2^2\right)$

$=\ln \left(2^2\right)-\ln \left(3\right)$

$\mathrm{Apply\:log\:rule}:\quad \log _c\left(a\right)-\log _c\left(b\right)=\log _c\left(\frac{a}{b}\right)$

$\ln \left(2^2\right)-\ln \left(3\right)=\ln \left(\frac{2^2}{3}\right)$

$=\ln \left(\dfrac{2^2}{3}\right)$

$2^2=4$

$=\ln \left(\dfrac{4}{3}\right)$

$=\dfrac{2\ln \left(2\right)}{\ln \left(\dfrac{4}{3}\right)}$

$x=\dfrac{2\ln \left(2\right)}{\ln \left(\dfrac{4}{3}\right)}$