How do you solve $2^{x} \cdot 5 = 10^{x}$ $2^x*5=10^x$ ?

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Answer 1 · 2016-06-30T00:53:32

If $A = B$ $A = B$ , then $log A = log B$ $logA = logB$ :

$log (2^{x} \times 5) = log (10^{x})$ $log(2^x xx 5) = log(10^x)$

Use the rule ${log}_{a} (\frac{n}{m}) = {log}_{a} n - {log}_{a} m$ $log_a(n/m) = log_an - log_am$ to simplify.

${log 2}^{x} + log 5 = {log 10}^{x}$ $log2^x + log5 = log10^x$

Use the rule ${log a}^{n} = n log a$ $loga^n = nloga$ to simplify further.

$x log 2 + log 5 = x log 10$ $xlog2 + log5 = xlog10$

$log 5 = x log 10 - x log 2$ $log5 = xlog10 - xlog2$

Factor out the $x$ $x$ .

$log 5 = x (log 10 - log 2)$ $log5 = x(log10 - log2)$

Put back the logs on the left side into quotient form to simplify.

$log 5 = x (log (\frac{10}{2}))$ $log5 = x(log(10/2))$

$log 5 = x (log 5)$ $log5 = x(log5)$

Isolate x.

$\frac{log 5}{log 5} = x$ $log5/log5 = x$

$1 = x$ $1 = x$

Checking in the original equation, we find that this solution works.

Hopefully this helps!

Answer 2 · 2016-06-30T00:56:28

$x = 1$ $x=1$ (using alternate solution method)

If $2^{x} \cdot 5 = 10^{x}$ $2^x*5=10^x$

$\Rightarrow 2^{x} \cdot 5 = {(2 \cdot 5)}^{x}$ $rArr 2^x*5 =(2*5)^x$

$\Rightarrow 2^{x} \cdot 5 = 2^{x} \cdot 5^{x}$ $rArr 2^x*5 = 2^x*5^x$

$\Rightarrow 5 = 5^{x}$ $rArr 5=5^x$

$\Rightarrow x = 1$ $rArr x=1$

How do you solve 2^x*5=10^x2x⋅5=10x2^x*5=10^x?