Exponential equation to solve for $x:\;\;$ $2^{5x-7} \cdot 5^{2x-1} =10^{x+1}$

Question

$$2^{5x-7} \cdot 5^{2x-1} =10^{x+1}$$

So here I tried taking logs of both sides but I could get further.

Answer 1

We have $$2^{5x−7}\cdot 5^{2x−1}=10^{x+1}$$

We can use exponent laws to rewrite this:

\begin{align}2^{5x−7}\cdot 5^{2x−1}&=10^{x+1}\\ \frac{2^{5x}}{2^7}\cdot \frac{5^{2x}}{5}&=10^{x+1}\tag{1}\\ \frac{2^{5x}}{2^7}\cdot \frac{5^{2x}}{5}&=10^x\cdot 10\tag{2}\\ \frac{2^{5x}\cdot 5^{2x}}{640}&=10^x\cdot 10\\ \frac{2^{5x}\cdot 5^{2x}}{10^x}&=6400\\ \frac{32^x\cdot 25^x}{10^x}&=6400\tag{3}\\ \frac{800^x}{10^x}&=6400\tag{4}\\ 80^x&=6400\tag{5}\\ x&=\log_{80}(6400)\\ x&=2\tag{6}\end{align}

Justification of each step:

$(1)$: $x^{a-b}=\dfrac{x^a}{x^b}$

$(2)$: $x^{a+b}=x^a\cdot x^b$

$(3)$: $x^{ab} = (x^a)^b$

$(4)$: $a^xb^x = (ab)^x$

$(5)$: $\dfrac{a^x}{b^x} = \left(\dfrac ab\right)^x$

$(6)$: We can do this just by looking at the line $80^x=6400$ and remembering that $8^2=64$ and $10^2=100$, therefore $80^2=8^2\times 10^2=6400$

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We could also take logs of both sides and solve as follows, however this is more complex:

\begin{align}2^{5x−7}\cdot 5^{2x−1}&=10^{x+1}\\ \ln\left(2^{5x−7}\cdot 5^{2x−1}\right)&=\ln\left(10^{x+1}\right)\\ \ln\left(2^{5x−7}\right)+\ln\left( 5^{2x−1}\right)&=\ln\left(10^{x+1}\right)\tag{1}\\ (5x-7)\ln(2)+(2x-1)\ln(5)&=(x+1)\ln(10)\tag{2}\\ 5\ln(2)x-7\ln(2)+2\ln(5)x-\ln(5)&=\ln(10)x+\ln(10)\\ (5\ln(2)+2\ln(5)-\ln(10))x&=\ln(10)+7\ln(2)+\ln(5)\\ x&=\frac{\ln(10)+7\ln(2)+\ln(5)}{5\ln(2)+2\ln(5)-\ln(10)}\\ x&=\frac{\ln(10)+\ln(2^7)+\ln(5)}{\ln(2^5+\ln(5^2)-\ln(10)}\tag{inverse of 2}\\ x&=\frac{\ln(10\times2^7\times5)}{\ln\left(\frac{2^5\times 5^2}{10}\right)}\tag{inverse of 1}\\ x&=\frac{\ln(6400)}{\ln(80)}\\ x\ln(80)&=\ln(6400)\\ \ln(80^x)&=\ln(6400)\tag{inverse of 2}\\ e^{\ln(80^x)}&=e^{\ln(6400)}\\ 80^x&=6400\tag{3}\\ x&=2\tag{4}\end{align}

Justification of each step:

$(1)$: $\ln(ab)=\ln(a)+\ln(b)$

$(2)$: $\ln(a^b)=b\ln(a)$

$(3)$: $e^{\ln a} = a$

$(4)$: Again, we can do this just by remembering that $8^2=64$ and $10^2=100$, therefore $80^2=8^2\times 10^2=6400$

Answer 2

We need to solve $$\frac{32^x\cdot25^x}{10^x}=2^7\cdot5^1\cdot10$$ or $$80^x=6400,$$ which gives $x=2$.

Answer 3

$$\ln(2^{5x-7}\cdot 5^{2x-1}) = \ln(10^{x+1})$$

$$\iff \ln(2^{5x-1}) + \ln(5^{2x-1}) = \ln(10^{x+1})$$

$$\iff (5x-1)\ln(2) + (2x-1)\ln(5) = (x+1)\ln(10).$$

I'll let you take it from here.

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I use the properties of logs:

• $\log_a(xy) =\log_a x + \log_a y$

• $\log_a(x^y) = y\cdot \log|a(x)$

I used $\log_e = \ln$.